Difference between revisions of "2016 MPFG Problem 18"
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<math>F(w) = F(w^2) = F(w^3) = F(w^4) = 2^3 = 8</math> | <math>F(w) = F(w^2) = F(w^3) = F(w^4) = 2^3 = 8</math> | ||
| − | <math>\frac{2^15+4\cdot8}{5} = 6560</math> | + | <math>\frac{2^{15}+4\cdot8}{5} = \boxed{6560}</math> |
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| + | ~cassphe | ||
Latest revision as of 10:48, 26 August 2025
Problem
Let 
. Say that a subset 
of 
is handy if the sum of all the elements of is a multiple of 
. For example, the empty set is handy (because its sum is 
) and itself is handy (because its sum is 
). Compute the number of handy subsets of .
Solution 1
We can use roots of unity to solve this problem.
~cassphe
